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Math Help - Vector space

  1. #1
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    Vector space

    I don't really understand some of this notation. I was wondering if someone could help me determine which of the following are subspaces and why they fail to be subspaces (if they are not).

    Let R[X] denote the vector space (over R) of polynomials with real coefficients. Which of the following are subspaces of R[X]?

    a) The set of odd polynomials.
    b) The set of polynomials of odd degree (We include the polynomial 0).
    c) {p(X) in R[X]: p(i*sqrt(3))=0}.
    d) {p(X) in R[X]: p(0)= i*sqrt(3)}.
    e) The set {(X^2+3)(p(X)): p(X) in R[X]}.
    f) The set {(X-i*sqrt(3))p(X): p(X) in R[X]}.
    g) The set of polynomials with rational coefficients.
    h) {p(X) in R[X]: p(i*sqrt(3)) is real}.


    Thank you so much for any help.
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  2. #2
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    Quote Originally Posted by thesummerofgeorge View Post
    I don't really understand some of this notation. I was wondering if someone could help me determine which of the following are subspaces and why they fail to be subspaces (if they are not).

    Let R[X] denote the vector space (over R) of polynomials with real coefficients. Which of the following are subspaces of R[X]?
    The fundamental theorem states this: if V is a vector space over the real field \mathbb{R} and some subset of V is closed under vector addition and scalar multiplication that is:
    \alpha+\beta \in V
    a\alpha \in V
    For all vectos in the subset and all real scalars then this subset is in fact a subspace.


    a) The set of odd polynomials.
    And odd polynomial, meaning
    P(-x)=-P(x)
    Thus,
    P(x)+Q(x)
    P(-x)+Q(-x)=-P(x)-Q(x)=-(P(x)+Q(x))
    Thus it is closed under vector addition.
    kP(x)
    kP(-x)=-kP(x)
    Thus, it is closer under scalar multiplication.
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  3. #3
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    Quote Originally Posted by thesummerofgeorge View Post
    I don't really understand some of this notation. I was wondering if someone could help me determine which of the following are subspaces and why they fail to be subspaces (if they are not).

    Let R[X] denote the vector space (over R) of polynomials with real coefficients. Which of the following are subspaces of R[X]?

    a) The set of odd polynomials.
    b) The set of polynomials of odd degree (We include the polynomial 0).
    Not a sumspace as x^3, -x^3+x^2 are both in the set, but their sum
    is not.

    c) {p(X) in R[X]: p(i*sqrt(3))=0}.
    Is a subspace, as if P and Q are in this set, for any a and b in R, we
    have:

    aP(i*sqrt(3)) + bQ(i*sqrt(3))=a*0+b*0=0.

    (also the set is not empty, but if it were it would still be a subspace)

    d) {p(X) in R[X]: p(0)= i*sqrt(3)}.
    If the set is non-empty then this is not, using the notation of part c) above, and a != -b

    aP(0)+bQ(0)=(a+b)*i*sqrt(3) != i*sqrt(3)

    however the set is empty, as any polynomial in x for which p(0)=i*sqrt(3)
    is not a polynomial with real coeficients, Hence this set is the null set and so
    is a sub-space


    RonL
    Last edited by CaptainBlack; January 14th 2007 at 11:20 AM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by thesummerofgeorge View Post
    e) The set {(X^2+3)(p(X)): p(X) in R[X]}.
    First the set is not empty. Let P(x)=(x^2+3)p(x), for some p(x) in R[x],
    and Q(x)=(x^2+3)q(x), for some q(x) in R[x], then these are both in R[x],
    and for any a and b in R:

    aP(x)+bQ(x)=(x^2+3)(p(x)+q(x))

    hence is in the set, so the set is a subspace.

    f) The set {(X-i*sqrt(3))p(X): p(X) in R[X]}.
    Is not a subspace as the elements of this set are not polynomials
    in x with real coeficients.

    g) The set of polynomials with rational coefficients.
    h) {p(X) in R[X]: p(i*sqrt(3)) is real}.
    Both of these are but I will leave it to you or some other helper
    to show they are.

    RonL
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